Question

Show that square of an odd positive integer is of the form 8q + 1, for some positive integer q.NCERT Class XMathematics - Exemplar ProblemsChapter _QUESTION PAPER SET II

Answer 1

let x be da number
x=8x
x²=(8x)²
x²=64x²
x²=8(8x²)
x²=8q [ let 8x² be q]
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x=(8q+1)
x²=(8q+1)²
x²=64q²+16q+1
x²=8q(8q+2)+1
x²=(8q+1) [ let 8q+2 be q]
hence proved 

Answer 2

Step-by-step explanation:

Hey friends !!

[ Note :- I am taking q as some integer . ]

Let 'a' be the any positive integer.

Then, b = 8 .

Using Euclid's division lemma :-

0 ≤ r < b => 0 ≤ r < 8 .

•°• The possible values of r is 0, 1, 2, 3, 4, 5, 6, 7.

▶ Question said Square of odd positive integer , then r = 1, 3, 5, 7 .

→ Taking r = 1 .

a = bm + r .

= (8q + 1)² .

= 64m² + 16m + 1

= 8( 8m²+ 2m ) + 1 .

= 8q + 1 . [ Where q = 8m² + 2m ]

→ Taking r = 3 .

a = bq + r .

= (8q + 3)² .

= 64m² + 48m + 9 = 64m² + 48m + 8 + 1 .

= 8( 8m²+ 6m + 1 ) + 1 .

= 8q + 1 . [ Where q = 8m² + 6m + 1 ]

→ Taking r = 5 .

a = (8q + 5)² .

= 64m² + 80m + 25 = 64m² + 80m + 24 + 1 .

= 8( 8m²+ 10m + 3 ) + 1 .

= 8q + 1 . [ Where q = 8m² + 10m + 3 ]

→ Taking r = 7 .

a = ( 8q + 7 )² .

= 64m² + 112m + 49 = 64m² + 112m + 48 + 1 .

= 8( 8m²+ 14m + 6 ) + 1 .

= 8q + 1 . [ Where q = 8m² + 14m + 6 ] .

Hence, the square of any odd positive integer is of the form 8q + 1 .

✓✓ Proved ✓✓

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